Page 62 - A First Course In Stochastic Models

POISSON ARRIVALS SEE TIME AVERAGES 53

and the long-run average sojourn time per accepted customer. The formulas (2.3.1),
(2.3.2) and (2.3.4) need only slight modification:

L q = λ(1 − P rej )W q and L = λ(1 − P rej )W, (2.3.6)
the long-run average number of customers in service
= λ(1 − P rej )E(S). (2.3.7)

Heuristically, these formulas follow by applying the money principle (2.3.3) and
taking only the accepted customers as paying customers.

2.4 POISSON ARRIVALS SEE TIME AVERAGES

In the analysis of queueing (and other) problems, one sometimes needs the long-
run fraction of time the system is in a given state and sometimes needs the long-
run fraction of arrivals who find the system in a given state. These averages can
often be related to each other, but in general they are not equal to each other. To
illustrate that the two averages are in general not equal to each other, suppose that
customers arrive at a service facility according to a deterministic process in which
the interarrival times are 1 minute. If the service of each customer is uniformly
distributed between 1 minute and 3 minute, then the long-run fraction of time the
4 4
system is empty equals 1 , whereas the long-run fraction of arrivals finding the
2
system empty equals 1. However the two averages would have been the same if
the arrival process of customers had been a Poisson process. As a prelude to the
generally valid property that Poisson arrivals see time averages, we first analyse
two specific problems by the renewal-reward theorem.

Example 2.4.1 A manufacturing problem
Suppose that jobs arrive at a workstation according to a Poisson process with rate
λ. The workstation has no buffer to store temporarily arriving jobs. An arriving job
is accepted only when the workstation is idle, and is lost otherwise. The processing
times of the jobs are independent random variables having a common probability
distribution with finite mean β. What is the long-run fraction of time the workstation
is busy and what is the long-run fraction of jobs that are lost?
These two questions are easily answered by using the renewal-reward theorem.
Let us define the following random variables. For any t ≥ 0, let

1 if the workstation is busy at time t,
I (t) =
0 otherwise.
Also, for any n = 1, 2, . . . , let

1 if the workstation is busy just prior to the nth arrival,
I n =
0 otherwise.